Suppose I know the mass and volume of an object, as well as its exact shape and density distribution as a function of space (x,y,z). Without submerging it to determine the amount of water displaced, how can I determine the buoyant force? Is there a way to determine the amount of displaced water based on knowing the density of the object as well as its shape?

Ultimately, I'd like to use this information to predict whether a particular design I have made will sink or float, but I need to know the buoyant force in order to do this. How can I predict this force a priori without experimentation?

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    $\begingroup$ Well, if you say you know the volume... $\endgroup$ – Solar Mike Mar 27 '18 at 17:46
  • $\begingroup$ @SolarMike: the volume of the object itself, not the volume of the water displaced by the object... $\endgroup$ – Paul Mar 27 '18 at 18:00
  • $\begingroup$ So what do you think the immersed volume is then? $\endgroup$ – Solar Mike Mar 27 '18 at 18:05
  • $\begingroup$ Or, to put it another way - what is the max immersed volume and will that volume of displaced water weigh more than the object? $\endgroup$ – Solar Mike Mar 27 '18 at 18:08
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    $\begingroup$ To be clear, can this object's average density be lower than water's? That is, can it float? Because if it can't, then there's obviously no difference between the object's volume and its displaced volume since the whole thing will sink. $\endgroup$ – Wasabi Mar 27 '18 at 18:13

Don't refer to the volume as the volume of material. Use volume as the outer dimensions of the object.

The volume of the water will be equal to the mass of the object (it if floats). Just divide the mass of the object by the density of water for displaced water. That will also be the exact volume of your object that is below the water line.

It does not matter to the water if the object is hollow or not. If the mass / volume of the object is greater than that of water it will sink.


with the clarifications that for a sunken object the volume of displaced fluid is the volume of the object, and for a floating object on a liquid, the weight of the displaced liquid is the weight of the object

  • $\begingroup$ So, the effective volume of the object (not just the volume of material) is used to compute the density of the object), and if it is less than the density of water, it floats? Also, what about the buoyant force? Is it always equal to the weight of the object regardless of whether or not it floats? $\endgroup$ – Paul Mar 28 '18 at 23:18
  • $\begingroup$ @Paul, if an object floats the force of buoyancy is its weight obviously. If it sinks the force of buoyancy is its volume in liters multiplied by kilograms. Just in case of boats and ships the force changes drastically if they sink, the volume will then be the volume of material, not the voolume of the vessel, which will be a much diminished buoyancy. $\endgroup$ – kamran Mar 29 '18 at 0:18
  • $\begingroup$ @Paul I feel like I answered the question. Good luck. $\endgroup$ – paparazzo Mar 29 '18 at 4:14

The buoyancy is equal to weight of water of displaced volume of the object.

If it's heavier than water it will sink and as long as you know the volume you know the buoyancy!

But if the object is not heavier than water, it will float. A part of the object will submerge, displacing (in the static case) exactly enough water to equalize the water's buoyant force with the object's weight. The level of water at which this occurs is named the draft line.

We need to find the object's CG, and its center of geometry.

$$ CG = (\sum \Delta v.\rho_v)/v $$ and center of geometry is a analytical process depending on its shape or combination of shapes and surfaces.

The axis of float is vertical from CG to center of geometry and draft volume is such that $$\Sigma \Delta v.\rho\space \text{submerged volume of water} = \text{weight of object}$$ ** Edit**

The submerged volume is a given and is equall to the weight of the object. However to calculate the draft line one needs to compile the geometey of the object as a function of the vertical axis say Z.

Then by equating the volume integral of submerged volume by the weight we get the Z of draft-line. Z is the upper limit of integration.

This process for many nicely behaving volumes is rather easy for instance for a submerged cone the volume is $$ V=2/3 Z. \pi .sin(theta)^2$$ and $\theta$ is half the cone angle.

$$Z = 2/3 . \pi.sin(\theta)^2/V $$

  • $\begingroup$ So don't you have two unknowns there... $\endgroup$ – Solar Mike Mar 27 '18 at 19:06
  • $\begingroup$ @SolarMike, What do you mean two unkown variables. $\endgroup$ – kamran Mar 27 '18 at 19:26
  • $\begingroup$ The vol submerged and vol of water... ie exactly that which the OP wants to find out... $\endgroup$ – Solar Mike Mar 27 '18 at 19:56
  • $\begingroup$ @Solar Mike, I have modified my answer to address your question. This is the subject of Naval and Marine engineering. Not only the object can be non- homogeneous, but the liquid too can be stratified. $\endgroup$ – kamran Mar 27 '18 at 22:55
  • $\begingroup$ @kamran Then density of water only changes 0.01% in 2000 meters of depth. $\endgroup$ – paparazzo Mar 28 '18 at 16:04

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